# Angular momentum in QM

Izbitzer
Hello,

I'm trying to solve a problem dealing with finding the probability of measuring certain values of $$L^2$$ for a particle.

The particle is on a sphere and is in the state $$\Psi (\theta , \phi) = Ne^{\\cos{\theta }}$$.

I don't know quite how to start, I guess I have to decompose the wave function in eigenfunctions for $$L^2$$, and then find the corresponding eigenvalues, and form that find the probability of measuring that particular eigenvalue, but like I said, I don't really know where to start.

Does anybody have any pointers?

Thanks!

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Homework Helper
Nope, you have to do you wrote. That is try to write your wavefunction as a linear combination of angular momentum eigenstates.

Daniel.

Izbitzer

I've worked some more, and I think I've made some progress The eigenfunctions of $$\^{L}^2$$ are spherical harmonics, but since the wave function is independent of $$\phi$$, m = 0, i.e $$\Psi(\theta,\phi) = \sum_{l} c_{l,0}Y_l^0$$.
The eigenvalues are $$l(l+1)\hbar$$. So I just see which value of $$l$$ corresponds with the values I'm supposed to find the probability of. The probability is calculated with: $$|c_{l}|^2 = |<\Psi|Y_l^0>|^2$$. This is where I'm stuck at the moment. The integrals are really complicated and I can't find them in any books, is there any easier way to calculate this?

Thanks!

Last edited:
Homework Helper
Gold Member
Izbitzer said:

I've worked some more, and I think I've made some progress The eigenfunctions of $$\^{L}^2$$ are spherical harmonics, but since the wave function is independent of $$\phi$$, m = 0, i.e $$\Psi(\theta,\phi) = \sum_{l} c_{l,0}Y_l^0$$.
The eigenvalues are $$l(l+1)\hbar$$. So I just see which value of $$l$$ corresponds with the values I'm supposed to find the probability of. The probability is calculated with: $$|c_{l}|^2 = |<\Psi|Y_l^0>|^2$$. This is where I'm stuck at the moment. The integrals are really complicated and I can't find them in any books, is there any easier way to calculate this?

Thanks!
A suggestion: I don't recall the exact form of the$Y^l_0$, but aren't they simple to write in terms of $(cos \theta)^n$ ? Then you simply have to Taylor expand $e^{cos \theta)}$. The coefficients will simply be the usual 1/n!. If you have a closed form expression for the $Y^l_0 (cos (\theta)$, then you can find the coeeficients c_l without doing a single integral.

Patrick

Izbitzer
nrqed said:
A suggestion: I don't recall the exact form of the$Y^l_0$, but aren't they simple to write in terms of $(cos \theta)^n$ ? Then you simply have to Taylor expand $e^{cos \theta)}$. The coefficients will simply be the usual 1/n!. If you have a closed form expression for the $Y^l_0 (cos (\theta)$, then you can find the coeeficients c_l without doing a single integral.

Patrick

That probably would have worked, but I managed to solve the damn integrals before I read your reply All it took was some (a lot of) integration by parts, and it turned out ok. Now I can finally rest. Thanks for taking the time guys!